If $\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right] \ldots \ldots \ldots . .\left[\begin{array}{cc}1 & n-1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 78 \\ 0 & 1\end{array}\right]$, then the inverse of $\left[\begin{array}{cc}1 & n \\ 0 & 1\end{array}\right]$ is:

(D) $\left[\begin{array}{cc}1 & -13 \\ 0 & 1\end{array}\right]$

(A) $\left[\begin{array}{cc}1 & -12 \\ 0 & 1\end{array}\right]$

(B) $\left[\begin{array}{cc}1 & 0 \\ 13 & 1\end{array}\right]$

(C) $\left[\begin{array}{cc}1 & 0 \\ 12 & 1\end{array}\right]$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Matrices

[1011] [1021] [1031] \dots\dots\dots .. [10 n - 1 1] = [10781]

, then the inverse of

[10 n 1]

is:

Answer: (D)

Multiplying the shear matrices

[10 k 1]

simply adds the top-right entries, so the product has top-right

1 + 2 + \dots + (n - 1) = \frac{n ( n - 1 )}{2}

, set equal to

78

. Then

n (n - 1) = 156

, giving

n = 13

. The matrix

[10131]

has inverse

[10 - 13 1]

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option